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Chapter 30: Problem 66

In the RC high-pass filter shown in the figure, \(R=10.0 \mathrm{k} \Omega\) and \(C=0.0470 \mu \mathrm{F}\) What is the 3.00 -dB frequency of this circuit (where \(\mathrm{dB}\) means basically the same for electric current as it did for sound in Chapter 16 )? That is, at what frequency does the ratio of output voltage to input voltage satisfy \(20 \log \left(V_{\text {out }} / V_{\text {in }}\right)=-3.00 ?\)

Short Answer

Expert verified
Answer: The 3.00-dB frequency of this RC high-pass filter circuit is approximately 723.6 Hz.

Step by step solution

01

Find the voltage transfer function

The voltage transfer function for an RC high-pass filter is given by the formula: \(H(f) = \frac {V_{out}}{V_{in}} = \frac {1}{\sqrt{1 + (\frac{1}{2\pi fRC})^2}}\). We will plug in the given values of R and C into this formula to find the transfer function.
02

Set \(20 \log \left(V_{\text {out }} / V_{\text{in }}\right)\) equal to -3.00

Given the equation, \(20 \log \left(V_{\text {out }} / V_{\text{in }}\right)=-3.00\), we will substitute \(V_{\text{out}}/V_{\text{in}}\) with the transfer function we found in Step 1: \(20 \log \left( \frac {1}{\sqrt{1 + (\frac{1}{2\pi fRC})^2}} \right) = -3.00\).
03

Solve for frequency f

To solve for the frequency, f, follow these steps: 1. Divide both sides by 20: \(\log \left( \frac {1}{\sqrt{1 + (\frac{1}{2\pi fRC})^2}} \right) = -0.15\). 2. Use the properties of logarithms to remove the log and square root: \(\frac {1}{\sqrt{1 + (\frac{1}{2\pi fRC})^2}} = 10^{-0.15}\) and then \({(10^{-0.15})^2 = \frac{1}{1 + (\frac{1}{2\pi fRC})^2}\) 3. Rearrange the equation to solve for \(f\): \((\frac{1}{2\pi fRC})^2 = \frac{1 - (10^{-0.15})^2}{(10^{-0.15})^2}\). 4. Take the square root of both sides: \(\frac{1}{2\pi fRC} = \sqrt{\frac{1 - (10^{-0.15})^2}{(10^{-0.15})^2}}\). 5. Rearrange to solve for f: \(f = \frac{1}{2\pi RC \sqrt{\frac{1 - (10^{-0.15})^2}{(10^{-0.15})^2}}}\).
04

Calculate the 3.00-dB frequency

Now, plug in the given values of R and C into the equation to find the 3.00-dB frequency: \(f = \frac{1}{2\pi (10.0 \times 10^3 \Omega)(0.0470 \times 10^{-6} F) \sqrt{\frac{1 - (10^{-0.15})^2}{(10^{-0.15})^2}}}\) After calculating the above expression, we get: \(f \approx 723.6 Hz\) Therefore, the 3.00-dB frequency of this RC high-pass filter circuit is approximately 723.6 Hz.

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Most popular questions from this chapter

An LC circuit consists of a capacitor, \(C=2.50 \mu \mathrm{F},\) and an inductor, \(L=4.0 \mathrm{mH}\). The capacitor is fully charged using a battery and then connected to the inductor. An oscilloscope is used to measure the frequency of the oscillations in the circuit. Next, the circuit is opened, and a resistor, \(R\), is inserted in series with the inductor and the capacitor. The capacitor is again fully charged using the same battery and then connected to the circuit. The angular frequency of the damped oscillations in the RLC circuit is found to be \(20 \%\) less than the angular frequency of the oscillations in the LC circuit. a) Determine the resistance of the resistor. b) How long after the capacitor is reconnected in the circuit will the amplitude of the damped current through the circuit be \(50 \%\) of the initial amplitude? c) How many complete damped oscillations will have occurred in that time?

Why is RMS power specified for an AC circuit, not average power?

An inductor with inductance \(L=47.0 \mathrm{mH}\) is connected to an AC power source having a peak value of \(12.0 \mathrm{~V}\) and \(f=1000 . \mathrm{Hz} .\) Find the reactance of the inductor and the maximum current in the circuit.

A capacitor with capacitance \(C=5.00 \cdot 10^{-6} \mathrm{~F}\) is connected to an AC power source having a peak value of \(10.0 \mathrm{~V}\) and \(f=100 . \mathrm{Hz} .\) Find the reactance of the capacitor and the maximum current in the circuit.

A series circuit contains a \(100.0-\Omega\) resistor, a \(0.500-\mathrm{H}\) inductor, a 0.400 - \(\mu\) F capacitor, and a time-varying source of emf providing \(40.0 \mathrm{~V}\). a) What is the resonant angular frequency of the circuit? b) What current will flow through the circuit at the resonant frequency?
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